At 101.3 kPa and 298 K, which element could this diagram represent

PLEASE PLEASE HELP ME!!!A 450 ml gas sample has a pressure of 1.25 atm at 65 °C. What is the temperature, in °C, at which the ga

Arlecino [84]

Answer:

89°C

Explanation:

Combined Gas Law (P₁V₁)/T₁ = (P₂V₂)/T₂

(1.25 atm)(450 mL)/(65°C) = (0.89 atm)(865 mL)/T₂

8.653846154 = 769.85/T₂

T₂ = 769.85/8.653846154

T₂ = 88.96044444 = 89°C

6 0

3 years ago

A solution contains 0.036 M Cu2+ and 0.044 M Fe2+. A solution containing sulfide ions is added to selectively precipitate one of

Ratling [72]

Answer:

The precipitate is CuS.

Sulfide will precipitate at [S2-]= 3.61*10^-35 M

Explanation:

Step 1: Data given

The solution contains 0.036 M Cu2+ and 0.044 M Fe2+

Ksp (CuS) = 1.3 × 10-36

Ksp (FeS) = 6.3 × 10-18

Step 2: Calculate precipitate

CuS → Cu^2+ + S^2- Ksp= 1.3*10^-36

FeS → Fe^2+ + S^2- Ksp= 6.3*10^-18

Calculate the minimum of amount needed to form precipitates:

Q=Ksp

For copper we have: Ksp=[Cu2+]*[S2-]

Ksp (CuS) = 1.3*10^-36 = 0.036M *[S2-]

[S2-]= 3.61*10^-35 M

For Iron we have: Ksp=[Fe2+]*[S2-]

Ksp(FeS) = 6.3*10^-18 = 0.044M*[S2-]

[S2-]= 1.43*10^-16 M

CuS will form precipitates before FeS., because only 3.61*10^-35 M Sulfur Ions are needed for CuS. For FeS we need 1.43*10^-16 M Sulfur Ions which is much larger.

The precipitate is CuS.

Sulfide will precipitate at [S2-]= 3.61*10^-35 M

3 0

3 years ago

Plz help i’m gonna cry i’m in class rn

iragen [17]

Answer:

if ur gonna cry then just dont cry its simple logic guyss!!!!!!!!!!!!

Explanation:

3 0

3 years ago

Suppose 2.8 moles of methane are allowed to react with 5 moles of oxygen.

Ronch [10]

Answer : The limiting reagent is $O_2$

Solution : Given,

Moles of methane = 2.8 moles

Moles of $O_2$ = 5 moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the balanced reaction we conclude that

As, 2 mole of $O_2$ react with 1 mole of $CH_4$

So, 5 moles of $O_2$ react with $\frac{5}{2}=2.5$ moles of $CH_4$

From this we conclude that, $CH_4$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Hence, the limiting reagent is $O_2$

8 0

3 years ago

Iron (III) oxide and hydrogen react to form iron and water, like this: Fe 03(s)+3H9)2Fe(s)+3HO) At a certain temperature, a chem

belka [17]

The question is incomplete, here is the complete question:

Iron (III) oxide and hydrogen react to form iron and water, like this:

$Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)$

At a certain temperature, a chemist finds that a 8.9 L reaction vessel containing a mixture of iron(III) oxide, hydrogen, Iron, and water at equilibrium has the following composition.

Compound Amount

Fe₂O₃ 3.95 g

H₂ 4.77 g

Fe 4.38 g

H₂O 2.00 g

Calculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits.

Answer: The value of equilibrium constant for given equation is $1.0\times 10^{-4}$

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

For hydrogen gas:

Given mass of hydrogen gas = 4.77 g

Molar mass of hydrogen gas = 2 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

$\text{Molarity of hydrogen gas}=\frac{4.77}{2\times 8.9}\\\\\text{Molarity of hydrogen gas}=0.268M$

For water:

Given mass of water = 2.00 g

Molar mass of water = 18 g/mol

Volume of the solution = 8.9 L

Putting values in above expression, we get:

$\text{Molarity of water}=\frac{2.00}{18\times 8.9}\\\\\text{Molarity of water}=0.0125M$

For the given chemical equation:

$Fe_2O_3(s)+3H_2(g)\rightarrow 2Fe(s)+3H_2O(g)$

The expression of equilibrium constant for above equation follows:

$K_{eq}=\frac{[H_2O]^3}{[H_2]^3}$

Concentration of pure solids and pure liquids are taken as 1 in equilibrium constant expression.

Putting values in above expression, we get:

$K_{c}=\frac{(0.0125)^3}{(0.268)^3}\\\\K_{c}=1.0\times 10^{-4}$

Hence, the value of equilibrium constant for given equation is $1.0\times 10^{-4}$

6 0

4 years ago